Re: Is this possible? Residue computation leads to complexInfinity?
- To: mathgroup at smc.vnet.net
- Subject: [mg60725] Re: [mg60719] Is this possible? Residue computation leads to complexInfinity?
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Mon, 26 Sep 2005 01:36:04 -0400 (EDT)
- Reply-to: hanlonr at cox.net
- Sender: owner-wri-mathgroup at wolfram.com
Your Residue expression is unevaluated (Residue appears in output) since you
left out part of the Residue syntax. However, your argument was evaluated to
ComplexInfinity so z must already have had a value.
Clear[z];
FullSimplify[
Residue[(I*z)/(1-(2+4*a)*z^2+z^4),
{z,Sqrt[a+1]-Sqrt[a]}],a>0]
(I*(Sqrt[1 + 1/a] - 1))/(8*(-a + Sqrt[a^2 + a] - 1))
Bob Hanlon
>
> From: "kiki" <lunaliu3 at yahoo.com>
To: mathgroup at smc.vnet.net
> Date: 2005/09/25 Sun AM 02:36:33 EDT
> Subject: [mg60725] [mg60719] Is this possible? Residue computation leads to
complexInfinity?
>
> In[15]:=
> \!\(FullSimplify[Residue[\(\[ImaginaryI]\ z\)\/\(1 - \((2 + 4\ a)\)\ z\^2 +
> z\
> \^4\), \@\(a + 1\) - \@a], \ a > 0]\)
>
> Out[15]=
> \!\(Residue[ComplexInfinity, \(-\@a\) + \@\(1 + a\)]\)
>
>
>